Solve for $x$ : $6x^2 + 18x + 12 = 0$
Solution: Dividing both sides by $6$ gives: $ x^2 + {3}x + {2} = 0 $ The coefficient on the $x$ term is $3$ and the constant term is $2$ , so we need to find two numbers that add up to $3$ and multiply to $2$ The two numbers $2$ and $1$ satisfy both conditions: $ {2} + {1} = {3} $ $ {2} \times {1} = {2} $ $(x + {2}) (x + {1}) = 0$ Since the following equation is true we know that one or both quantities must equal zero. $(x + 2) (x + 1) = 0$ $x + 2 = 0$ or $x + 1 = 0$ Thus, $x = -2$ and $x = -1$ are the solutions.